Oxygen is contained in a pisto
Oxygen is contained in a piston-cylinder system initially at atemperature of 40 oC and a pressure of 150 kPa. Thepiston is 1 m in diameter and initially 20 cm above the base of thecylinder. Heat is added at constant pressure until the piston hasmoved 30 cm, so it’s now at 50 cm.
I found the mass of the system to be 0.2898 kg, and T2 =783K.
a) Using the table of properties of oxygen as an ideal gas,calculate the heat added to the system per unit mass, in kJ/kg
b) Repeat this calculation assuming that the specific heat as afunction of temperature is linear over this range of temperatures(ie, assuming the change in temperature is “small”)
c) Repeat this calculation without this assumption (ie, directintegration of empirical relation)
Answer:
Assume a piston-cylinder arrangement,with a piston diameter 
1m
area of the piston ( A)0.785* 12 =.785 m2
at an initial condition temperatureT1 273 + 40 313 K
Pressure P1= 150 kPa ,Volume V1= A * L1 = .785 * 0.2= 0.157m3 ( L1 = .2 m GIVEN)
Here in this system heat is added atconstant pressure so initial pressure P1 = finalpressure P2 = 150 kPa
as OXYGEN is assume be an ideal gasso , Cp = 1.005 KJ/Kg and Cv = .718 KJ/Kg (Oxygen is a diatomic gas so we assume this value )
mass of the system (m)= 0.2898 kg
final temp. T2 = 783 K
final volume V2 = A*L2 = .785 * 0.5 = 0.3925 m3
a.Heat added to the system per unit mass ( Q) =Cp*
T=1.005 * (783 – 313) = 472.35 KJ/Kg ( answer for a)
b.as the change of the temperature is small so we assume thatT = 0
As per 1st law of thermodynamics we know dQ = dU +dW
as T is equal to zero so dU = 0
so , dQ = dW = 
pdv = p (V2 – V1 ) = 150 *( 0.3925-0.157) = 35.325 kJ / Kg
heat added for the mass m is = 35.325 * 0.2898 = 10.237 KJ (answer for b)
c.assuming that the specific heat as a function of temperature islinear over this range of temperatures so Cp= a + bT (a& b are constant )
for initial condition for the table of O2Cp 
0.92 KJ/Kg
so , 0.92 = a + 313b ( equation 1)
for final condition for the O2 tableCp1.046 KJ/Kg
so , 1.046= a + 783b ( equation 2)
now from this 2 equation we get the value of a =0.833 and b= 2.766*10-4
so the equation become Cp = 0.833 + ( 2.766*10-4)T
SO heat added to this system by this process isper unit mass (dQ) = CpT
total heat added ( Q ) = 
Cp T = (0.833 + (2.766*10-4)T) T
=(0.833( T 2 – T1
) + ( ( 2.766*10-4)/ 2 ) * (T22 -T12))
put T1 = 313 andT2 = 783 then we get
Q = 0.833 ( 783 – 313 ) + ( 1.383 *10-4) ( 7832 – 3132 )
= 391.51 + 71.24 = 462.75 KJ/Kg
Total heat added for the system mass (m) = 462.75 * 0.2898 =134.1 KJ ( answer for C)
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
